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JEE MainChemistryIonic Equilibrium

40 mL of 0.2 M NH ₄ OH is titrated with 20 mL of 0.1 M HCl . The pH of the resulting solution will be _ _ _ _ 10⁻² . (Nearest integer) (Given : p K_b( NH ₄ OH ) = 4.74 , 3 = 0.48 )

Correct answer

974

Step-by-step solution

Initial millimoles of NH ₄ OH = 40 0.2 = 8 mmol Millimoles of HCl added = 20 0.1 = 2 mmol The reaction is: NH ₄ OH + HCl NH ₄ Cl + H ₂ O Since HCl is the limiting reagent, 2 mmol of NH ₄ Cl is formed and the remaining NH ₄ OH is 8 - 2 = 6 mmol . The resulting solution is a basic buffer. pOH = p K_b + ( [ Salt ] [ Base ] ) pOH = 4.74 + ( 2 6 ) = 4.74 + ( 1 3 ) pOH = 4.74 - 3 = 4.74 - 0.48 = 4.26 pH = 14 - pOH = 14 - 4.26 = 9.74 Thus, pH = 974 10⁻² . Answer: 974

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